I am trying to prove the following:
⊢ (□p ∨ □q) → □(p ∨ q)
However, I think that I am lacking the knowledge of a tautology in classical logic that would help me prove this.
I tried something, but it seems too contrived to work and I'm thinking there's an easier solution.
1. p → (p ∨ q) tautology 2. □(p → (p ∨ q)) necessity axiom 3. □p → □(p ∨ q) distribution axiom * 4. q → (p ∨ q) tautology 5. □(q → (p ∨ q)) necessity axiom 6. □q → □(p ∨ q) distribution axiom **
Now, I wanted to use the following tautology from classical logic:
⊢ (A → C) → (B → C) → ((A ∨ B) → C)
Where I would substitute A=□p, B=□q, C=□(p ∨ q) which I think would get me the result I want. However, I am unable to get □p and □q alone.
Even more, if I'm to use this in my test, I'd need to prove the tautology via natural deduction first and then use it. Since this is the case, I thought that there had to be another, more easier way to do it.
Could someone help?